Oxygen is more electro negative yet in the redox reaction

$$\ce{2PbO(s) -> 2Pb(s) + O2(g)}$$

my text book says lead is reduced and oxygen is oxidized. How can this be? It doesn't make sense to me. Oxygen is usually $2-$ while lead is usually $2+$.

  • $\begingroup$ Oxidation is by definition a loss of electrone(s). $\ce{2O^{2-} - 4e -> O2^0}$ is an oxidation of oxygen. I honestly don't see what confuses you so much. $\endgroup$
    – andselisk
    Jul 15, 2017 at 6:01
  • 2
    $\begingroup$ I agree with the OP that oxidation is a confusing word, especially if applied to the element oxygen. It would be better to say, in the example given, that the lead(+2) cation is reduced to elemental lead (0) and the oxide (-2) anion is oxidized to elemental oxygen (0). By analogy: converting chloride ion to chlorine gas is clearly oxidation, so the same is true for oxygen. $\endgroup$
    – iad22agp
    Jul 15, 2017 at 10:43

1 Answer 1


Your confusion stems from that fact that you are trying to look at what reactions are favorable instead of the reaction as written.

Oxygen is readily reduced because it is a strong oxidant. You can check any reduction half reaction with oxygen on the left side. Chances are that the reduction potential will be fairly positive.

In the reaction you have written, oxide is being oxidized to oxygen. Nowhere (including your textbook) do we claim that this reaction is favorable or spontaneous in the forward direction. In fact, if you consult a table of free energies of formation (example), you would find an entry like:

$$\ce{PbO (s, red)}, \Delta G^{\varnothing}_{f}=-189.3\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$ or $$\ce{PbO (s, yellow)}, \Delta G^{\varnothing}_{f}=-188.5\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$

These values indicate that indeed, the reverse reaction is the one that's favorable, the one where oxygen serves as an oxidizing agent and is reduced.

However, that is not the reaction as written. The notation of chemistry allows us to describe reactions that are hypothetical or unfavorable.


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